A volume of 80.0 mL of aqueous potassium hydroxide (KOH) is titrated against a standard solution of sulfuric acid (H2SO4). What was the molarity of the KOH solution if 12.7 mL of 1.50 M H2SO4 was needed

0.478 M

Explanation:

Let's consider the neutralization reaction between KOH and H₂SO₄.

2 KOH + H₂SO₄ → K₂SO₄ + 2 H₂O

12.7 mL of 1.50 M H₂SO₄ react. The reacting moles of H₂SO₄ are:

0.0127 L × 1.50 mol/L = 0.0191 mol

The molar ratio of KOH to H₂SO₄ is 2:1. The reacting moles of KOH are 2 × 0.0191 mol = 0.0382 mol

0.0382 moles of KOH are in 80.0 mL. The molarity of KOH is:

M = 0.0382 mol/0.0800 L = 0.478 M

it takes on an electronical energy